$$(x-5)(x+2)(x-5)+3\cdot(-3)\cdot(-4)-3\cdot(-4)\cdot4 = 0$$
Answer
$$ \begin{matrix}x_1 = -3.23781 & x_2 = 5.61891+3.1327i & x_3 = 5.61891-3.1327i \end{matrix} $$
Explanation
$$ \begin{aligned} (x-5)(x+2)(x-5)+3\cdot(-3)\cdot(-4)-3\cdot(-4)\cdot4 &= 0&& \text{simplify left side} \\[1 em](x-5)(x+2)(x-5)+36+48 &= 0&& \\[1 em](x^2+2x-5x-10)(x-5)+36+48 &= 0&& \\[1 em](x^2-3x-10)(x-5)+36+48 &= 0&& \\[1 em]x^3-5x^2-3x^2+15x-10x+50+36+48 &= 0&& \\[1 em]x^3-8x^2+5x+134 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
This page was created using
Polynomial Equations Solver